Question:medium
If \( a_1, a_2, a_3, \dots \) are in A.P., then the value of \( \begin{vmatrix} a_1 & a_2 & 1 \\ a_2 & a_3 & 1 \\ a_3 & a_4 & 1 \end{vmatrix} \) is equal to:
If \( a_1, a_2, a_3, \dots \) are in A.P., then the value of \( \begin{vmatrix} a_1 & a_2 & 1 \\ a_2 & a_3 & 1 \\ a_3 & a_4 & 1 \end{vmatrix} \) is equal to:
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Determinants involving A.P. terms often resolve to zero because the "constant increase" creates linear dependency between rows.
Updated On: May 6, 2026
- \( a_4 - a_1 \)
- \( \frac{a_1 + a_4}{2} \)
- \( 1 \)
- \( \frac{a_2 + a_3}{2} \)
- \( 0 \)
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The Correct Option is
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